Abstract

Well-known Banach space results (e.g., due to J. Koliha and Y. Katznelson/L. Tzafriri), which relate conditions on the spectrum of a bounded operator to the operator norm convergence of certain sequences of operators generated by , are extended to the class of quojection Fréchet spaces. These results are then applied to establish various mean ergodic theorems for continuous operators acting in such Fréchet spaces and which belong to certain operator ideals, for example, compact, weakly compact, and Montel.

1. Introduction

Given a Banach space and a continuous linear operator on , there are various classical results which relate conditions on the spectrum of with the operator norm convergence of certain sequences of operators generated by . For instance, if , with denoting the operator norm, (even in the weak operator topology suffices), then necessarily , where , [1, p. 709, Lemma 1]. The stronger condition is equivalent to the requirement that both and hold [2]. An alternate condition, namely, that is a convergent sequence relative to the operator norm, is equivalent to the requirement that the three conditions , the range is closed in for some , and are satisfied [3]. Here with being the boundary of . Such results as above are often related to the uniform mean ergodicity of , meaning that the sequence of averages of is operator norm convergent. For instance, if and , then is uniformly mean ergodic [4, p. 90, Theorem 2.7]. Or if , then is uniformly mean ergodic if and only if is closed [5].

Our first aim is to extend results of the above kind to the class of all Fréchet spaces referred to as prequojections; this is achieved in Section 3. The extension to the class of all Fréchet spaces is not possible; see Proposition 17 below and [6, Example 3.11], for instance. We point out that a classical result of Katznelson and Tzafriri stating, for any Banach-space-operator satisfying , that if and only if [7], is also extended to prequojection Fréchet spaces; see Theorem 20.

Our second aim is inspired by well-known applications of the above mentioned Banach space results to determine the uniform mean ergodicity of operators which satisfy and belong to certain operator ideals, such as the compact or weakly compact operators; see, for example, [1, Ch. VIII, 8], [4, Ch. 2, 2.2], and [8, Theorem 6.1], where can even be quasi-compact. An extension of such a mean ergodic result to the class of quasi-precompact operators acting in various locally convex Hausdorff spaces is presented in [9]. For prequojection Fréchet spaces, this result is further extended to the (genuinely) larger class of quasi-Montel operators; see Proposition 32, Remark 33, and Theorem 35. A mean ergodic theorem for Cesàro bounded, weakly compact operators (and also reflexive operators) in a certain class of locally convex spaces (which includes all Fréchet spaces), is also presented; see Proposition 23 and Remark 24(ii).

2. Preliminaries and Spectra of Operators

Let be a lcHs and a system of continuous seminorms determining the topology of . The strong operator topology in the space of all continuous linear operators from into itself (from into another lcHs we write ) is determined by the family of seminorms , for , for each and , in which case we write . Denote by the collection of all bounded subsets of . The topology of uniform convergence on bounded sets is defined in via the seminorms , for , for each and ; in this case we write . For a Banach space, is the operator norm topology in . If is countable and is complete, then is called a Fréchet space. The identity operator on a lcHs is denoted by .

By we denote equipped with its weak topology , where is the topological dual space of . The strong topology in (resp. ) is denoted by (resp. ) and we write (resp. ); see [10, IV, Ch. 23] for the definition. The strong dual space of is denoted simply by . By we denote equipped with its weak-star topology . Given , its dual operator is defined by for all , . It is known that and , [11, p. 134].

For a Fréchet space and , the resolvent set of consists of all such that exists in . Then is called the spectrum of . The point spectrum consists of all such that is not injective. Unlike for Banach spaces, it may happen that . For example, let be the Fréchet space equipped with the lc-topology determined via the seminorms , where , for . Then the unit left shift operator , for , belongs to and, for every , the element is an eigenvector corresponding to .

For a Fréchet space , the natural imbedding is an isomorphism of onto the closed subspace of . Moreover, we always have
that is, is an extension of .

The following result will be required in the sequel. Since the proof is standard we omit it. The polar of a set is denoted by .

Lemma 1. Let be a Fréchet space.(i)Let be a fundamental, increasing sequence which determines the lc-topology of . For each define on via . Then is a fundamental, increasing sequence which determines the lc-topology of .(ii)Let be a fundamental, increasing sequence which determines the lc-topology of . For each , let denote the Minkowski functional (in ) of the bipolar of . Then is a fundamental, increasing sequence which determines the lc-topology of . Moreover, for each , we have
for each and . In particular, ; that is, the restriction of to coincides with , for each .

For Banach spaces the following fact is well-known.

Lemma 2. Let be a lcHs and be an equicontinuous sequence. Then also is equicontinuous.

Proof. Let . Then as is equicontinuous. So, for all and , we have with
As the seminorms generate the lc-topology of , the previous inequality shows that is equicontinuous.Since is equicontinuous and the lc-topology of is generated by the polars of bounded subsets of , the same argument as above yields that is equicontinuous.

Lemma 3. Let be a Fréchet space and . Then is an isomorphism of onto itself if and only if is an isomorphism of onto itself.

Proof. If is an isomorphism of onto itself, then there exists with . It follows that and so . Accordingly, and . Thus, exists in and ; that is, is an isomorphism of onto itself.Conversely, suppose that is an isomorphism of onto itself. Since is an extension of (i.e., ), we see that is one-to-one. Moreover, since is a closed subspace of (as is a complete, barrelled lcHs), it follows that is closed. It remains to show that . But, if , then there is such that for all . Hence, ; this is a contradiction because the surjectivity of implies that is necessarily one-to-one.

Corollary 4. Let be a Fréchet space and . Then and . Moreover,
that is, the restriction of to the closed subspace of coincides with . Briefly, .

A Fréchet space is always a projective limit of continuous linear operators , for , with each a Banach space. If and can be chosen such that each is surjective and is isomorphic to the projective limit , then is called a quojection [12, Section 5]. Banach spaces and countable products of Banach spaces are quojections. Actually, every quojection is the quotient of a countable product of Banach spaces [13]. In [14] Moscatelli gave the first examples of quojections which are not isomorphic to countable products of Banach spaces. Concrete examples of quojection Fréchet spaces are , the spaces , with , and for , with any open set, all of which are isomorphic to countable products of Banach spaces. The spaces of continuous functions , with a -compact, completely regular topological space, endowed with the compact open topology, are also quojections. Domański exhibited a completely regular topological space such that the Fréchet space is a quojection which is not isomorphic to a complemented subspace of a product of Banach spaces, [15, Theorem]. A Fréchet space admits a continuous norm if and only if contains no isomorphic copy of [16, Theorem ]. On the other hand, a quojection admits a continuous norm if and only if it is a Banach space [12, Proposition 3]. So, a quojection is either a Banach space or contains an isomorphic copy of , necessarily complemented, [16, Theorem ]. Also [17] is relevant.

Lemma 5. Let be a quojection Fréchet space and . Suppose that , with a Banach space (having norm ) and linking maps which are surjective for all , and suppose, for each , that there exists satisfying
where , , denotes the canonical projection of onto (i.e., ). Then
Moreover,
If, in addition, for every , the resolvent operator satisfies
then .

Proof. For the containments (6) and (7) we refer to [18, Lemma 5.1].Suppose now that (8) holds for each . To establish the desired equality, let . Then is surjective. Fix . Since is surjective, it is routine to check from the identity that also is surjective (with the identity operator). To verify is injective suppose that for some , in which case for some . Accordingly,
shows that . It then follows from (8) that ; that is, . Since , we have . Hence, is injective. This establishes that . Accordingly, as desired.

Lemma 6. Let be a quojection Fréchet space and . Suppose that , with a Banach space (having norm ) and linking maps which are surjective for all , and suppose, for each , that there exists satisfying
where , , denotes the canonical projection of onto (i.e., ). Then the following statements are equivalent. (i)The limit - exists in .(ii)For each , the limit - exists in . In this case, the operators and , for , satisfy
Moreover, (i) and (ii) remain equivalent if is replaced by .

Given any lcHs and , let us introduce the notation:
for the Cesàro means of . Then is called mean ergodic precisely when is a convergent sequence in . If happens to be convergent in , then will be called uniformly mean ergodic.

We always have the identities
and also (setting ) that
Some authors prefer to use in place of ; since
this leads to identical results.

Recall that is called power bounded if is an equicontinuous subset of .

The final result that we require (i.e., [18, Lemma 5.4]) is as follows.

Lemma 7. Let be a quojection Fréchet space and let operators and , for , be given which satisfy the assumptions of Lemma 5 (with , , denoting the canonical projection of onto and being the norm in the Banach space ). (i) is power bounded if and only if each , , is power bounded.(ii) is mean ergodic (resp., uniformly mean ergodic) if and only if each , , is mean ergodic (resp., uniformly mean ergodic).

3. Spectrum, Uniform Convergence, and Mean Ergodicity

A prequojection is a Fréchet space such that is a quojection. Every quojection is a prequojection. A prequojection is called nontrivial if it is not itself a quojection. It is known that is a prequojection if and only if is a strict (LB) space. An alternative characterization is that is a prequojecton if and only if has no Köthe nuclear quotient which admits a continuous norm; see [12, 19–21]. This implies that a quotient of a prequojection is again a prequojection. In particular, every complemented subspace of a prequojection is again a prequojection. The problem of the existence of nontrivial prequojections arose in a natural way in [12]; it has been solved, in the positive sense, in various papers [19, 22, 23]. All of these papers employ the same method, which consists in the construction of the dual of a prequojection, rather than the prequojection itself, which is often difficult to describe (see the survey paper [24] for further information). However, in [25] an alternative method for constructing prequojections is presented which has the advantage of being direct. For an example of a concrete space (i.e., a space of continuous functions on a suitable topological space), which is a nontrivial prequojection, see [26].

In this section we extend to prequojection Fréchet spaces some well-known results from the Banach setting which connect various conditions on the spectrum , of a continuous linear operator , to the operator norm convergence of certain sequences of operators generated by . Such results have well-known consequences for the uniform mean ergodicity of .

We begin with a construction for quojection Fréchet spaces which is needed in the sequel.

Let be a quojection Fréchet space and be any fundamental, increasing sequence of seminorms generating the lc-topology of . For each , set and endow with the quotient lc-topology. Denote by the corresponding canonical (surjective) quotient map and define the quotient topology on via the increasing sequence of seminorms on by
for each . Then
Moreover,
which implies that is a norm on . As noted above, since is a quojection Fréchet space and every quotient space (of such a Fréchet space) with a continuous norm is necessarily Banach [12, Proposition 3], it follows that for each there exists such that the norm generates the lc-topology of . Moreover, it is possible to choose for all . Thus, is isomorphic to the projective limit of the sequence of Banach spaces with respect to the continuous, surjective linking maps defined by
This particular construction will be used on various occasions in the sequel, where will always denote the closed unit ball of , for . The so-constructed Banach space norm of will always be denoted by , for .

The following result is classical in Banach spaces [1, p. 709, Lemma 1].

Proposition 8. Let be a quojection Fréchet space and satisfy -. Then .In case is a prequojection Fréchet space and -, the inclusion is again valid.

Proof. We have the following two cases.Case (I) ( is a quojection). Let be a fundamental, increasing sequence of seminorms generating the lc-topology of . Since in as and is a Fréchet space, the sequence is equicontinuous. So, for each there exists such that
there is no loss in generality by assuming that can be chosen.Define on by , for . Then (20) ensures that is also a fundamental, increasing sequence of seminorms generating the lc-topology of . Moreover,
We now apply the construction (16)–(19) to the sequence of seminorms to yield the corresponding sequence of Banach spaces and the quotient maps , for ; recall that , for .Fix . Define the operator via
Then is a well-defined, continuous linear operator from into . Indeed, suppose for some ; that is, , so that . This, together with (21), yields . Since , it follows that , and hence, by (22) that . Therefore, . This means that is well defined. Clearly, is also linear. Moreover, (17), (21), and (22) imply that
for all and with . Taking the infimum with respect to , it follows that
Since generates the quotient topology of , (24) ensures the continuity of . Moreover, it follows from (22) that
The surjectivity and the continuity of together with (25) imply that -. Indeed, fix any . By the surjectivity of there exists such that . By (25) it follows that , for . Moreover, as by assumption. So, the continuity of yields that in the Banach space . We can then apply Lemma 1 in [1, p. 709] to obtain that .We have just shown that . Moreover, the operators and satisfy (22). So, we can apply Lemma 5 which yields ; that is, .Case (II). ( is a prequojection and -). Observe that and are barrelled and, hence, quasi-barrelled as is a Fréchet space and is the strong dual of a prequojection Fréchet space. Since and , the condition - implies that - (see [27, Lemma 2.6] or [28, Lemma 2.1]). On the other hand, is a quojection Fréchet space. So, it follows from Case (I) that . Finally, Corollary 4 ensures that and so .

Remark 9. For a power-bounded operator it is always the case that - and so, whenever is a prequojection Fréchet space, it follows from Proposition 8 that .

For operators in Banach spaces, the following result is due to Koliha [2].

Theorem 10. Let be a prequojection Fréchet space and . The following assertions are equivalent. (i)-.(ii)The series converges in .(iii)- and .Moreover, if one (hence, all) of the above conditions holds, then is an isomorphism of onto with inverse and the series converging in .

Proof. We have the following two cases.Case (I) ( is a quojection). (i)(ii). The assumption - implies that -. So, we can proceed as in the proof of Proposition 8 to obtain that in such a way that, for every , there exists in satisfying . Then also , for every . So, Lemma 6 implies that - for all . Thus, by [2, Theorem 2.1] the series converges in , for each . With , for , it follows again from Lemma 6 that the series converges in .(ii)(iii). The assumption clearly implies -. So, as in the proof of (i)(ii), we may assume that in such a way that, for every , there exists in satisfying . Then also , for every . Since converges in and is a quojection, the series also converges in for all ; see Lemma 6. By [2, Theorem 2.1] we have that and so , for all . Accordingly, since for all , Lemma 5 yields ; that is, .(iii)(i). Since , for every , the operator is invertible, that is, bijective with . On the other hand, - for every as - and . So, by Theorem 4.1 in [29] (see also Theorem 3.5 of [6]) we can conclude that
Let be a fundamental, increasing sequence of seminorms generating the lc-topology of . Arguing as in the proof of Proposition 8 (and adopting the notation from there) we conclude that (20) is satisfied. Define on by , for . Then again (21) is satisfied and, for each , there exists a continuous linear operator satisfying both (22) and (24). Moreover, it follows from (22) that
Fix and consider the sequences and in given by and , for . Then the operator satisfies for all . Moreover, (26) implies that in . Since all the assumptions of Lemma 3.4 in [6] are satisfied with , , and , we can proceed as in the proof of that result to conclude, for every , that the operator is invertible in (hence, also is invertible); that is, .By the arbitrariness of , we have that , for all . So, there exists such that . It follows that
and, hence, that . Because of (27), with , it follows from Lemma 6 (with ) that -.Case (II) ( is a prequojection). As noted before and are barrelled with and .(i)(ii). If in for , then an argument as for Case (II) in the proof of Proposition 8 shows that in for . Since is a quojection Fréchet space, we can apply (i)(ii) of Case (I) above to conclude that the series converges in . Then also converges in as and is a closed subspace of .(ii)(iii). If converges in , then converges in ; see [27, Lemma 2.6] or [28, Lemma 2.1]. Since is a quojection Fréchet space, we can apply (ii)(iii) of Case (I) above to conclude that (the condition - clearly follows from the assumption). So, by Corollary 4.(iii)(i). As already noted (cf. proof of Case (II) in Proposition 8) and are barrelled (hence, quasi-barrelled) and -. By Corollary 4, and so by assumption. Since is a quojection Fréchet space, we can apply Case (I) to conclude that -. So, also - as and is a closed subspace of .Finally, suppose that one (hence, all) of the above conditions holds. Then the series converges in and so in for . But, for every , we have
and so, for , we can conclude that with convergence of the series in . In a similar way one shows that , with the series again converging in .

Remark 11. In the proof of (iii)(i) in Case (I) above, if , then it follows that . But, this is not the case in general as the following example shows.Let be a Banach space and let be an increasing sequence with . Consider the quojection Fréchet space (endowed with the product topology) and the operator on defined by , for . It is easy to show that and that is even power bounded. Moreover, . Indeed, for a fixed , if , then ; that is, for all . Since , it follows that for all and so . On the other hand, if , then belongs to and . Hence, is bijective and so . Moreover, fix any and set for every . Then for every . Thus, each is an eigenvalue of .Now, suppose that for some . Then . But for , and hence, there is such that . This contradiction as is an eigenvalue for .

If is uniformly mean ergodic, then (14) implies that -. With an extra condition the converse is also valid.

Corollary 12. Let be a prequojection Fréchet space and . If - and , then is uniformly mean ergodic.

Proof. Since , the operator is bijective and so the space is closed in . By [6, Theorem 3.5], is uniformly mean ergodic. In particular, as , we have that in for .

Remark 13. Let be a prequojection Fréchet space and let satisfy -. If , then the proof of Corollary 12 shows that is uniformly mean ergodic with -. On the other hand, if (a stronger condition than ), then Theorem 10 implies that - and hence again - follows [30, Remark 3.1]. However, the stronger conclusion that - does not follow from Corollary 12 in general. Indeed, let be any Banach space (even finite dimensional). Then every power of belongs to the set and so is power bounded. This implies that -. Since , surely and so, by Corollary 12, it follows that -. However, for every we have and so does not converge to zero. This does not contradict Theorem 10 as is not included in .

Remark 14. Let be a prequojection Fréchet space and . We observe the following.(i)Proposition 8 and (14) yield that if is uniformly mean ergodic, then - and .(ii)Suppose that -. If , then is uniformly mean ergodic and - (cf. Remark 13).For Banach spaces the next result is due to Mbekhta and Zemànek [3]. Recall that .

Theorem 15. Let be a prequojection Fréchet space and . The following statements are equivalent. (i) is convergent in .(ii)-, the linear space is closed in for some and .(iii)- and is closed for some .

Proof. (i)(ii). If converges in to , say, then is uniformly mean ergodic with ergodic projection equal to [30, Remark 3.1]. Moreover, as is necessarily equicontinuous, it follows that -. Hence, by Theorem 3.5 and Remark 3.6 of [6] the space is closed for every . Moreover, by Proposition 8 we have . To establish the remaining condition we distinguish two cases.(a) is a quojection. Let be any fundamental, increasing sequence of seminorms generating the lc-topology of . By equicontinuity of , for each , there exists such that
Define , for each , by , for . Then (30) ensures that is also a fundamental, increasing sequence of seminorms generating the lc-topology of . Moreover, it is routine to check (using also that for each ) that
With (31) in place of (21), we can argue as in the proof of Proposition 8 to deduce that and that, for every , there exist operators and in satisfying and . Hence, for every . Since also -, it follows from Lemma 6 (with and ) that -, for each . By [3, Corollaire 3] we have that for every . This implies that . Indeed, if , then for every we have and so ; that is, . As for every , an appeal to Lemma 5 yields that .(b) is a prequojection. As noted before, and are barrelled (hence, quasi-barrelled) with and . Hence, - implies that -; see [27, Lemma 2.6] or [28, Lemma 2.1]. Since is a quojection Fréchet space, we can apply the result from case (a) to conclude that and so ; see Corollary 4.(ii)(i). The assumptions - and the space being closed for some imply that is uniformly mean ergodic [6, Theorem 3.4 and Remark 3.6]. In particular, is closed and
[6, Theorem 3.4]. Moreover, Proposition 8 implies that . It then follows from the assumption that either or .If , then necessarily and so, by (iii)(i) of Theorem 10, we have -.In the event that we have that and so (otherwise, is injective and from also surjective; that is, ). Define and . Then is a prequojection Fréchet space (being a quotient space of the prequojection ) which is -invariant and so . The claim is that
It follows from (32) that . Fix (so that ). If for some (i.e., ), then as . Hence, is injective. Next, let . Then there exists such that . Since with and (cf. (32)), it follows that ; that is, , with and . As and , this implies that and so with ; that is, is surjective. These facts show that . This establishes .Fix . Suppose that for some . Then with and (cf. (32)). It follows that with and . Arguing as in the previous paragraph, this implies that and . Since and , we can conclude that ; that is, is injective. Next, let . Then with and (cf. (32)). Since , the element exists. Moreover, with implies the existence of such that . It follows that satisfies . Hence, is also surjective and so . Accordingly, is proved. This establishes (33).Since and (33) is equivalent to , it follows that . Moreover, is a prequojection Fréchet space and in as (because - and on ). So, we can apply Theorem 10 to conclude that in as . On the other hand, on . These facts ensure that in because and on .(i)(iii). If converges to some in , then is uniformly mean ergodic with ergodic projection equal to [30, Remark 3.1]. Hence, by [6, Theorem 3.5 and Remark 3.6] the space is closed for every . Moreover, in as .(iii)(i). We first observe that
This identity (together with the fact that - implies for the averages that - [30, Remark 3.1]) yields -. But, - and so we can conclude that -. As also is closed for some , we can apply [6, Theorem 3.4 and Remark 3.6] to conclude that is uniformly mean ergodic and, in particular, that (32) is valid with being closed. We claim that this fact, together with the assumption that -, implies that converges in . To see this, note that on and so in as . On the other hand, the surjective operator lifts bounded sets via [10, Lemma 26.13] because and , both being prequojections, are quasinormable Fréchet spaces [24, Proposition 2.1], [21]; that is, for every there exists such that . So, for fixed (with corresponding set ) and (every is the restriction of some ), we have
where as by assumption. Set . The arbitrariness of and shows that in (after observing that is -invariant and so ). These facts ensure that in as .

Remark 16. In assertion (ii) of Theorem 15 the condition that “ is closed in for some ” can be replaced with the condition that “ is uniformly mean ergodic”; see [6, Theorem 3.5 and Remark 3.6].

Theorems 10 and 15 do not necessarily hold for operators acting in general Fréchet spaces.

Proposition 17. Let or and let be a Köthe matrix on such that is a Montel space with . Then there exists an operator such that in as and but is not closed for every .

Proof. By the proof of Proposition 3.1 in [6] there exists with for all such that the diagonal operator given by , for , is power bounded, uniformly mean ergodic and is dense but, not closed in . So, for every , also is dense but not closed in . To see this, note that the arguments in the proof of [6, Remark 3.6, (5)(4)] are valid for any operator satisfying - and acting in any Fréchet space. So, in the case that was closed for some , we could apply [6, Remark 3.6, (5)(4)] to conclude that is also closed; a contradiction. So .We claim that in as . Indeed, since is equicontinuous and convergence of a sequence in is equivalent to its convergence in (as is Montel), it suffices to show that in for each , where . But, this is immediate because , for all .It remains to show that . Set . Then . Let . Then . It is routine to check that, for a fixed , the element belongs to and satisfies . This means that the operator is surjective. On the other hand which follows from . Therefore, as is a Fréchet space, ; that is, . Since , it follows that .

Concerning the example in Proposition 17 we note that (i) of Theorem 10 holds but (iii) of Theorem 10 fails (as implies that ). Moreover, (i) of Theorem 15 holds (as -) but (ii) and (iii) of Theorem 15 fail (because is not closed in for every ). Of course, is not a prequojection.

A well-known result of Katznelson and Tzafriri states that a power bounded operator on a Banach space satisfies if and only if , [7, Theorem 1 and p. 317 Remark]. In order to extend this result to prequojection Fréchet spaces (see Theorem 20 below) we require the following notion.

Let be a Fréchet space and . A fundamental, increasing sequence which generates the lc-topology of is called contractively admissible if, for each